Local flatness for non-freely falling observers?

[center o bass]

The equivalence principle implies that it's always possible, by an appropriate coordinate transformation, to locally transform away the effects of gravity by going into a freely falling frame.

Furthermore the local flatness theorem implies that it's always possible to but the metric $g_{\mu \nu}$ into it's canonical form at any event $P$, i.e.
$g_{\mu \nu}(P) = \eta_{\mu \nu}$ such that the first derivatives of the metric components vanish. This is thus the mathematical expression of the equivalence principle and by the (stronger) Einstein equivalence principle, one can apply the laws of special relativity in these coordinates.

These considerations are often applied when considering redshift effects in general relativity. Consider an observer; at the even A he sees a photon fly by. By going into inertial coordinates for which the observer is at rest his four-velocity will be $U = (1,0,0,0)$ and the energy-momentum of the photon will be $P=(E,p_x, p_y, p_z)$ and thus the energy the observer would measure is given by

$$ E = - U \cdot P$$

and from this formula one can derive redshifts effects etc.

That is all nice, but doesn't the argument above require that the observer must be freely falling? If he is not freely falling, i.e. not on a geodesic, is it then also possible to construct local inertial coordinates for which U = (1,0,0,0)?

Often the formula above is applied to observers which are stationary in, say, Schwarzschild spacetime. But I would not suppose that these observers are freely falling, and if so is it then okay to use the above result for these observers?

If it is not possible to construct locally inertial coordinates for observers which is not freely falling, it seems like the result of such a computation would give the result of what a freely falling observer at the event A would measure and not necessarily what the observer of interest would measure.

 

[Bill_K]

Locally Minkowskian coordinates based on an arbitrary timelike curve are called Fermi coordinates.

 

[center o bass]

Locally Minkowskian coordinates based on an arbitrary timelike curve are called Fermi coordinates.

Is it always possible to transform the metric into Fermi coordinates? And is it physically allowed to apply SR in these coordinates when they are not freely falling?

If not, wouldn't it then be a problem that textbooks apply the formula above for non-freely falling observers?

 

[WannabeNewton]

Hi center o bass! I'm not understanding your question. The fact that you can set up normal coordinates on some neighborhood of any point on a pseudo-Riemannian manifold $M$ has nothing at all to do with observers, it's just a property of the manifold (Normal coordinates just capture the intuitive notion that $T_p M$ is the "best linear approximation" of $M$ at $p$ by using the exponential map). I'm not seeing any link to observers.

 

[pervect]

Is it always possible to transform the metric into Fermi coordinates? And is it physically allowed to apply SR in these coordinates when they are not freely falling?

If not, wouldn't it then be a problem that textbooks apply the formula above for non-freely falling observers?

You can always transform the metric so that it's diag(-1,1,1,1) locally. This is just a variable substitution - a linear one, at that.

That's all you really need to do what you want. In these local coordinates, you can interpret changes of coordinates as changes of distance.

The Christoffel symbols will be non-zero as the time derivatives of the metric will be non-zero, but this doesn't affect doppler shifts. I'm not going to try to make a list of what is and isn't affected by Christoffel symbols, I am hoping you'll be able to figure it out with some thought.

The cannonical example for understanding the effects of non-zero Christoffel symbols is Einstein's elevator. It's worth convincing yourself that the Christoffel symbols in the elevator must be the acceleration of the elevator if you're not familiar with the idea.

Fermi normal coordinates are slightly more involved, but they are applicable to both curved space-times as well as flat, so they'll include GR and SR.

Do you understand tangent spaces, also known as "local Lorentz frames"? I"m guessing you might not. Unfortunately, it's getting hard to continue to talk about Fermi normal coordinates without making some assumptions about what you do and don't know about regarding tangent spaces.

 

[BruceW]

Locally Minkowskian coordinates based on an arbitrary timelike curve are called Fermi coordinates.

wikipedia seem to be saying that Fermi coordinates are only allowed on a geodesic. (i.e. only a freely-falling observer).

 

[Bill_K]

wikipedia seem to be saying that Fermi coordinates are only allowed on a geodesic. (i.e. only a freely-falling observer).

You're right, they do say that!

Here's a better http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&page=articlesu18.html, which defines them for an arbitrary timelike curve, and also goes into more detail.

 

[BruceW]

I think the idea is that if we define normal coordinates around the point, then if we move along a geodesic through that point, the Christoffel symbols vanish (as long as we are close to the point, and when we get further away, then generally Christoffel symbols become nonzero again).

So to answer the original question, yes the observer needs to be free-falling to experience local 'flatness'. And yes, in your textbook example, it seems they are really showing you what is observed by a free-falling observer in the Schwarzschild spacetime, not one which is stationary.

 

[BruceW]

You're right, they do say that!

Here's a better http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&page=articlesu18.html, which defines them for an arbitrary timelike curve, and also goes into more detail.

ah, I see. thanks, its a good link. it defines Fermi coordinates in a better way than wikipedia. They define it for any timelike curve, and note that the Christoffel symbols vanish when that curve is a geodesic.

edit: Also, your link says that we can define Fermi normal coordinates along an entire timelike curve, and if the curve is a geodesic, then these 'Fermi normal coordinates' enforce local flatness along the entire geodesic. That's pretty cool. But I guess it makes sense, since at any point along the geodesic, we can define a coord system that has vanishing Christoffel symbols, therefore we can just let our coordinates evolve in such a way that at every point along the geodesic, the coordinates are exactly those which cause the Christoffel symbols to vanish.

 

[WannabeNewton]

Again, I'm not seeing what local flatness has to do with observers. It is simply a local properly of $C^{\infty}$ curved manifolds. The OP seems to be referring to the equivalence principle when asking about local flatness applied to freely falling vs. accelerating observers but the confusion is largely due to the horrible phrasing that some sources use with words like "freely falling frame" etc. when stating the equivalence principle. If you have a tensor equation, say $\nabla^{a}T_{ab} = 0$, and $p\in M$, then you can always find a neighborhood $U$ of $p$ with coordinate functions $x^{\mu}: U \rightarrow \mathbb{R}$ such that $\nabla^{\mu}T_{\mu\nu}(p) = \partial^{\mu}T_{\mu\nu}(p) = 0$ when represented in these coordinates i.e. the non-gravitational laws of general relativity reduce to the laws of special relativity locally in space-time; I never understood why some sources use terms like "freely falling frame" when stating this part of the equivalence principle (this doesn't even have anything to do with reference frames-it is simply a matter of employing normal coordinates). It is again just the intuitive statement that $T_p M$, which is isomorphic to $\mathbb{R}^{4}$, approximates $M$ at $p$.

There is no contradiction with regards to freely falling or accelerating observers: if an observer is freely falling and sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will find that at that event, he/she is traveling on a straight line (moving uniformly) i.e. $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = 0$. If an accelerating observer sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will still measure proper acceleration at that event, given in the normal coordinates by $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = a^{\mu}(p)$.

 

[center o bass]

Again, I'm not seeing what local flatness has to do with observers. It is simply a local properly of $C^{\infty}$ curved manifolds. The OP seems to be referring to the equivalence principle when asking about local flatness applied to freely falling vs. accelerating observers but the confusion is largely due to the horrible phrasing that some sources use with words like "freely falling frame" etc. when stating the equivalence principle. If you have a tensor equation, say $\nabla^{a}T_{ab} = 0$, and $p\in M$, then you can always find a neighborhood $U$ of $p$ with coordinate functions $x^{\mu}: U \rightarrow \mathbb{R}$ such that $\nabla^{\mu}T_{\mu\nu}(p) = \partial^{\mu}T_{\mu\nu}(p) = 0$ when represented in these coordinates i.e. the non-gravitational laws of general relativity reduce to the laws of special relativity locally in space-time; I never understood why some sources use terms like "freely falling frame" when stating this part of the equivalence principle (this doesn't even have anything to do with reference frames-it is simply a matter of employing normal coordinates). It is again just the intuitive statement that $T_p M$, which is isomorphic to $\mathbb{R}^{4}$, approximates $M$ at $p$.

There is no contradiction with regards to freely falling or accelerating observers: if an observer is freely falling and sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will find that at that event, he/she is traveling on a straight line (moving uniformly) i.e. $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = 0$. If an accelerating observer sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will still measure proper acceleration at that event, given in the normal coordinates by $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = a^{\mu}(p)$.

You're probably right as to where my confusion lies. I've always recognized the local flatness of being a property of the manifold itself, but as you stated some authors conclude with sentences like: "since a freely falling frame and an inertial frame far away from gravitational fields are equivalent the laws of SR can be applied locally in the freely falling frame", giving the misconception that you have to be in a freely falling frame to use SR locally.

However the equivalence principle also expresses full equivalence with gravitation and acceleration locally, and since the laws of SR hold instantaneously in an accelerated frame they should also hold locally for any frame regardless of the motion of an observer stationary in that frame. Thus an accelerated observer (with 4-acceleration) could also write an equation like

$$P = (E, p_x, p_y, p_z).$$

Agree?

 

[WannabeNewton]

I'm not sure what your question is. Consider a particle with 4-momentum $p$ at an event $q$ on its worldline. Let $O$ be an observer with 4-velocity $u$ who is present at $q$ ($O$ may or may not be accelerating). Choose a local Lorentz frame $\{e_{(a)}\}$ for $O$ at $q$. In this frame, $u = e_{(0)}$ and $p = p^{a}e_{(a)}$ so that $-g_{q}(p,u) = -p^{a}g_{q}(e_{{(a)}},e_{0}) = -p^{a}\eta_{a0} = -p^{0} = E$. Why do we need normal coordinates at all?

 

[BruceW]

...when represented in these coordinates i.e. the non-gravitational laws of general relativity reduce to the laws of special relativity locally in space-time; I never understood why some sources use terms like "freely falling frame" when stating this part of the equivalence principle (this doesn't even have anything to do with reference frames-it is simply a matter of employing normal coordinates).

isn't a "freely falling frame" just the coordinate system you have described? (I always thought that 'coordinate system' and 'reference frame' were synonyms). But from what you are saying here, it seems that they are not?

I've always recognized the local flatness of being a property of the manifold itself, but as you stated some authors conclude with sentences like: "since a freely falling frame and an inertial frame far away from gravitational fields are equivalent the laws of SR can be applied locally in the freely falling frame", giving the misconception that you have to be in a freely falling frame to use SR locally.

You are correct that the manifold is always locally 'flat' in that we can choose the metric to be diagonal at any instant. (As someone already said). But we can't generally make the Christoffel symbols vanish, unless we have a freely falling frame. So we need a freely falling frame to be able to use SR locally.

However the equivalence principle also expresses full equivalence with gravitation and acceleration locally, and since the laws of SR hold instantaneously in an accelerated frame they should also hold locally for any frame regardless of the motion of an observer stationary in that frame.

If you are not traveling along a geodesic, then locally you will have fictitious forces that you cannot get rid of. When you have fictitious forces that you cannot get rid of, you do not have local SR. And so to be able to generally say that you have local SR, you need to be traveling along a geodesic.

p.s. I am relatively ;) new to GR, so I'm not 100% sure about what I'm saying.

 

[WannabeNewton]

isn't a "freely falling frame" just the coordinate system you have described? (I always thought that 'coordinate system' and 'reference frame' were synonyms). But from what you are saying here, it seems that they are not?

A coordinate system and a frame are not the same thing, although some sources use them synonymously which does lead to confusions. At any event on an observer's worldline, we can find an orthonormal basis for the tangent space at that event. This orthonormal basis represents a system of measuring rods and a clock held by the observer-this is what we call the frame; there is no apriori reason to think this should have anything to do with any kind of coordinate chart. We can use such a frame to define a coordinate system on some neighborhood of the event, but this is not the same thing as saying the frame is a coordinate system by itself.

See the following quote:

It is also important to distinguish "coordinates" and "frames". A system of coordinates is a set of labels attached to various spacetime points. All it tells you is that, say, (x,y,z,t) corresponds to that place over there at that time.

You are correct that the manifold is always locally 'flat' in that we can choose the metric to be diagonal at any instant. (As someone already said). But we can't generally make the Christoffel symbols vanish, unless we have a freely falling frame. So we need a freely falling frame to be able to use SR locally.

This relates to what I said above. First let me define what a normal neighborhood is. Let $M$ be a Riemannian manifold and let $p\in M$. There always exists a neighborhood $U\subseteq M$ of $p$ and a neighborhood $V\subseteq T_pM$ of $0\in T_pM$ such that $\exp_p:V\rightarrow U$ is a diffeomorphism. We call $U$ a normal neighborhood of $p$. Now, let $\{e_{(a)}\}$ be an orthonormal basis for $T_p M$. This defines an isomorphism $E:\mathbb{R}^{n}\rightarrow T_pM$ given by $E(x^{1},...,x^{n}) = x^{a}e_{(a)}$. Using $E$ and the exponential map on a normal neighborhood $U$ of $p$, we can define a coordinate map $\varphi = E^{-1}\circ \exp_p^{-1}:U\rightarrow \mathbb{R}^{n}$. It is possible to show that in these coordinates, $\Gamma ^{i}_{jk}(p) = 0$ identically. These coordinates are called Riemannian normal coordinates, or just normal coordinates.

So all I needed was an orthonormal basis for $T_p M$ and the fact that I can always find a normal neighborhood of $p$. If we have an accelerating observer and an event on his/her worldline, I can always find a momentarily comoving locally inertial frame (MCLIF) at that event i.e. find a freely falling observer whose worldline intersects the accelerating observer at the given event, and whose 4-velocity at that event coincides with that of the accelerating observer, and adopt the inertial frame of the freely falling observer at that event for the accelerating observer's use. What's stopping me from finding a normal neighborhood of this event and using the orthonormal basis of the MCLIF to employ Riemann normal coordinates as defined above? I'm personally failing to see why we cannot do this.

 

[BruceW]

interesting stuff. I hope to learn more about those 'morphisms in the future. It seems a much nicer way to talk about GR. about the topic: I see what you're saying. If we choose 1 event on the worldline, then we can always define a geodesic through it, who sees the world as locally SR. But we can't relate this (in a useful way) to the person who is not traveling along the geodesic. This is because the person and the geodesic only intersect at exactly that 1 event. In the neighbourhood around that event, they will be doing different things. Therefore, the person does not see the world as locally SR. For him to see the world as locally SR, he would have to stay on the geodesic in some small neighbourhood around the event. But he does not.

Alright, to be more specific, his velocity is the same as that defined on the geodesic at that event, right? But his acceleration will not be. So even though his velocity carries him in the same direction as the geodesic, his acceleration will pull him away from this path. So I guess that to second order, he does not see the world as locally SR. He only sees the world as locally SR to first order.

for example, if he is measuring the path of a beam of light. If he measures the velocity of the beam of light, he gets the usual SR result. But if he measures the acceleration of the beam of light, he gets a result which does not match what he would expect from local SR.

And conversely, if a person traveling along a geodesic measures the acceleration of a beam of light, then he does get the result that agrees with local SR. This is because he is traveling along the geodesic, so he can use SR locally to get correct predictions.

 

[George Jones]

There two different types of normal coordinates being discussed in this thread, Fermi (by Bill) and Riemann (by WannabeNewton), and this might be causing some confusion.

Reimann normal coordinates. Consider an p as origin and some "nearby" point q. Pick an orthonormal frame of (tangent) vectors at p. Find the unique geodesic form p to q that has curve parameter 0 at p and 1 at q. At p, find the tangent vector to this curve. The Riemann normal coordinates of q are the components of this vector with respect to the chosen orthonormal frame at p. This establishes a coordinate system in an appropriate neighbourhood of q.

Fermi normal coordinates. Pick an event p on an observer's (accelerated or not) worldline, and choose an orthonormal frame at p for which e_0 is the observer's 4-velocity at p. Fermi transport the triad of spatial vectors all along the worldline. Consider a point q that is "nearby" the worldline, but not necessarily "nearby" p. From q, find the unique geodesic that orthogonally intersects the worldline (at p' say), and that has curve parameter 1 at q. At p', find the tangent vector to this curve. The Fermi normal spatial coordinates of q are the components of this vector with respect to the spatial triad at p' that was Fermi-transported from p. The Fermi normal time coordinate of q is difference in proper time between p and p'. This establishes a coordinate system in an appropriate neighbourhood tube along the *entire* worldline of the observer. The observer is *always* at the spatial origin of Fermi normal coordinates.

 

[WannabeNewton]

Fermi normal coordinates.

Hi George. I scrapped my earlier post because I have a direct question. Now, if we happen to use Fermi Normal coordinates along the worldline of an accelerating observer, the Christoffel symbols don't vanish identically along the entire worldline in these coordinates so at any given event on the worldline, the accelerating observer would not be able to, in these coordinates, write down the non gravitational laws of GR in SR form.

However, what's confusing me is, why can't we choose some arbitrary event on the accelerating worldline and employ Riemann normal coordinates in a neighborhood of this event so that the Christoffel symbols do vanish identically at this event in these coordinates, making non-gravitational tensor laws reduce to ones in SR at this event? Is there something preventing us from doing this? It makes intuitive sense to me that even an accelerating observer should be able to use SR locally, via this coordinate system, since the fact that space-time is locally Minkowskian is a property of the manifold and not dependent on whether the observer in the local region is accelerating or not.

Why then, if my intuition is not horribly flawed, do some books (e.g. Stephani, MTW) say that a "locally flat coordinate system" (i.e. Riemann normal coordinates) about an event is the coordinate system set up exclusively by a freely falling observer at that event, as opposed to any time-like observer at that event? I'm just not getting why only a freely falling observer can setup these locally flat coordinates about events on his/her worldline. I don't see where in the construction of Riemann normal coordinates this requirement is imposed.

As far as I can see, the only difference between this and Fermi Normal coordinates (other than the fact that in Fermi Normal coordinates the Christoffel symbols won't vanish in general) is that the latter is defined all along the accelerating worldline and we have a preset transport law for the local Lorentz frame erected at some initial event (the Fermi Walker transport) whereas in the former case we would need to keep (in principle) reconstructing Riemann normal coordinates at each consecutive event. Could you clarify this? Thanks George (and sorry for all the questions, it's my fault for learning exclusively from Schutz and Wald :p)!

 

[pervect]

If you are not traveling along a geodesic, then locally you will have fictitious forces that you cannot get rid of.

I agree

When you have fictitious forces that you cannot get rid of, you do not have local SR.

I disagree. This is a common misconception. You can handle accelerating observers, and even accelerating frames just fine in SR, by which I mean without using Einstein's field equations.

The methods used typically aren't taught in introductory SR courses, but you don't need any new theoretical assumptions to handle fictitious forces.

See for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html "Can special relativity handle accelerations."

It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames. It is claimed that general relativity is required because special relativity only applies to inertial frames. This is not true. Special relativity treats accelerating frames differently from inertial frames but can still deal with them. Accelerating objects can be dealt with without even calling upon accelerating frames.

And so to be able to generally say that you have local SR, you need to be traveling along a geodesic.

p.s. I am relatively ;) new to GR, so I'm not 100% sure about what I'm saying.

OK, hope this helps

 

[PeterDonis]

However, what's confusing me is, why can't we choose some arbitrary event on the accelerating worldline and employ Riemann normal coordinates in a neighborhood of this event so that the Christoffel symbols do vanish identically at this event in these coordinates, making non-gravitational tensor laws reduce to ones in SR at this event?

You can; but the coordinate chart you construct this way will be *different* than Fermi normal coordinates for that accelerating worldline.

Why then, if my intuition is not horribly flawed, do some books (e.g. Stephani, MTW) say that a "locally flat coordinate system" (i.e. Riemann normal coordinates) about an event is the coordinate system set up exclusively by a freely falling observer at that event, as opposed to any time-like observer at that event?

Do they really say that? Can you give some specific quotes that give you this impression? I'm not familiar with Stephani, but in MTW, I don't get the sense that they are saying that Riemann normal coordinates can *only* be used by freely falling observers.

However, there is a sense in which Riemann normal coordinates give freely falling observers a "special" place: the worldlines of freely falling observers are straight lines in Riemann normal coordinates. See below.

As far as I can see, the only difference between this and Fermi Normal coordinates (other than the fact that in Fermi Normal coordinates the Christoffel symbols won't vanish in general) is that the latter is defined all along the accelerating worldline and we have a preset transport law for the local Lorentz frame erected at some initial event (the Fermi Walker transport) whereas in the former case we would need to keep (in principle) reconstructing Riemann normal coordinates at each consecutive event.

This is true, but there's another difference, which I just alluded to. Let's take a concrete example: the worldline of an observer "hovering" at constant r in Schwarzschild coordinates. This observer is accelerated, so we can set up Fermi normal coordinates centered on his worldline. In these coordinates, the observer's worldline is a straight line.

Now pick an event E on that observer's worldline, and consider the worldline of a freely falling observer who just comes to rest for an instant at radial coordinate r at that event E. In the above Fermi normal coordinates (let's suppose we pick event E as the origin of those coordinates, to make it simple), the free-faller's worldline will be curved (I think it will look something like the way an inertial observer's worldline in Minkowski spacetime looks in Rindler coordinates). But if we set up Riemann normal coordinates centered on event E, in those coordinates the free-faller's worldline is a straight line, and the "hovering" observer's worldline will be curved (I think it will be a hyperbola, as a Rindler observer's worldline is in Minkowski coordinates in flat spacetime).

 

[PeterDonis]

Here's a better http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&page=articlesu18.html, which defines them for an arbitrary timelike curve, and also goes into more detail.

This reference also appears to indicate that, if the timelike curve is not a geodesic, the Fermi normal coordinates on that curve will *not* look like Minkowski coordinates. That may be the confusion underyling the Wikipedia article: what Wikipedia should have said is that Fermi normal coordinates can be set up for any timelike curve, but the metric will only take the Minkowski form at events on that curve if the curve is a geodesic.

 

[PAllen]

This reference also appears to indicate that, if the timelike curve is not a geodesic, the Fermi normal coordinates on that curve will *not* look like Minkowski coordinates. That may be the confusion underyling the Wikipedia article: what Wikipedia should have said is that Fermi normal coordinates can be set up for any timelike curve, but the metric will only take the Minkowski form at events on that curve if the curve is a geodesic.

What was really surprising to me when I first worked with Fermi-Normal coordinates for non-inertial world lines was that they achieve the metric taking Minkowski form on the whole defining world line; it is only the connection which doesn't vanish on this world line; and the connection then directly encodes proper acceleration (or even spin, if you allow for it using a sufficiently general formulation of FN coordinates). Compare to RN coordinates where both metric and connections are made to vanish at the chosen origin event.

 

[PeterDonis]

What was really surprising to me when I first worked with Fermi-Normal coordinates for non-inertial world lines was that they achieve the metric taking Minkowski form on the whole defining world line

Yes, on reviewing pp. 329-332 of MTW, I see that I mis-stated things in post #20. The Living Reviews article that Bill_K linked to didn't seem as clear to me as the MTW discussion. I believe that what I said in post #19 is still correct; however, there is an interesting point about that that I think is worth raising here.

Consider Rindler coordinates in flat spacetime. These coordinates make the worldlines of accelerated observers look like straight lines. Question: are Rindler coordinates equivalent to Fermi-Normal coordinates for one of those accelerated observers?

 

[WannabeNewton]

Do they really say that? Can you give some specific quotes that give you this impression? I'm not familiar with Stephani, but in MTW, I don't get the sense that they are saying that Riemann normal coordinates can *only* be used by freely falling observers.

Here are some quotes, perhaps you can interpret them better than I, because I might have misread them: "But geodesics are paths of particles in the gravitational field (as we have made plausible and shall later prove); that is, for the observers just mentioned there exists (locally) no gravitational field: the locally flat coordinate system is the system of a freely falling
observer at the point in question of the space-time." (Stephani page 115) - It's that last part that confuses me, I don't see what they mean by that.

On second thought, MTW's paragraph is actually not as confusing as the one in Stephani. All they say is "Pick an event $P_0$ and a set of basis vectors $\{e_{\alpha}(P_0)\}$ to be used there by an inertial observer" (MTW page 285). I suppose for an accelerating observer, at $P_0$, I can interpret this to mean we take a momentarily comoving locally inertial observer at $P_0$ with the associated frame $\{e_{\alpha}(P_0)\}$?

This is true, but there's another difference, which I just alluded to. Let's take a concrete example: the worldline of an observer "hovering" at constant r in Schwarzschild coordinates. This observer is accelerated, so we can set up Fermi normal coordinates centered on his worldline. In these coordinates, the observer's worldline is a straight line.

Now pick an event E on that observer's worldline, and consider the worldline of a freely falling observer who just comes to rest for an instant at radial coordinate r at that event E. In the above Fermi normal coordinates (let's suppose we pick event E as the origin of those coordinates, to make it simple), the free-faller's worldline will be curved (I think it will look something like the way an inertial observer's worldline in Minkowski spacetime looks in Rindler coordinates). But if we set up Riemann normal coordinates centered on event E, in those coordinates the free-faller's worldline is a straight line, and the "hovering" observer's worldline will be curved (I think it will be a hyperbola, as a Rindler observer's worldline is in Minkowski coordinates in flat spacetime).

Ok so let's assume we have a stationary observer $O$ in Schwarzschild space-time at some radius $r > 2GM$. $O$ must be following an orbit of $\xi^{a}$ and as such has a non-vanishing acceleration of magnitude $a = (\nabla^{b}\ln V)(\nabla_{b}\ln V) = \frac{1}{\sqrt{1 - \frac{2m}{r}}}\frac{m}{r^{2}}$ (say due to the rocket thrusters $O$ is using to hover in place at $r$). Now let's say we employ Riemann normal coordinates about some event $E$ on $O$'s worldline and also have a freely falling observer $O'$ comoving with $O$ at $E$ (so comes to rest for an instant at $E$ as you said yes?). In Riemann normal coordinates $\{x^{\mu}\}$, the equations of motion for $O$ at $E$ become $\frac{\mathrm{d} ^{2}x^{\mu}_{O}}{\mathrm{d} \tau^{2}}(E) = a^{\mu}_{O}(E)$ with the RHS non-vanishing in Riemann normal coordinates (I haven't exactly computed it however) whereas for $O'$ we have $\frac{\mathrm{d} ^{2}x^{\mu}_{O'}}{\mathrm{d} \tau^{2}}(E) = 0$ in Riemann normal coordinates.

On the other hand, if we employ Fermi Normal coordinates $\{x^{\mu'}\}$ using $O$'s worldline (which work everywhere on $O$'s worldline but we just care about $E$) you are saying that the equations of motion for the respective observers in these coordinates, at $E$, will look like $\frac{\mathrm{d} ^{2}x^{\mu'}_{O}}{\mathrm{d} \tau^{2}}(E) = 0$ and $\frac{\mathrm{d} ^{2}x^{\mu'}_{O'}}{\mathrm{d} \tau^{2}}(E) = a^{\mu'}_{O'}(E)$ with the RHS non-vanishing?

So are Fermi Normal coordinates sensitive to the worldline we choose them to be along? What I mean by this is, even though both $O$ and $O'$ intersected at $E$ and were comoving there, since we chose Fermi Normal coordinates along $O$'s worldline, this essentially acted as a "rest frame" of $O$ so that $O$ would look like he was traveling in a straight line in these coordinates whereas $O'$ would not, even at $E$, because even though they are comoving at $E$, overall $O'$ has a different motion from that of $O'$. If we instead chose Fermi Normal coordinates along $O'$, and considered the same scenario at $E$, would now $O'$ look like he was traveling in a straight line whereas $O'$ would look like he has a curved motion, even at $E$?

In this second case wouldn't Fermi normal coordinates evaluated at $E$ just be Riemann normal coordinates at $E$? So in a sense, can we consider Riemann normal coordinates to describe unequivocally the "rest frame" of freely falling observers, whereas the notion of Fermi Normal coordinates describing the "rest frame" of something, in general, is sensitive to who's worldline we use them on i.e. if we use them along the stationary observer then Fermi normal coordinates will be describing his "rest frame" but if we use them along a different worldline then the stationary observer's trajectory will, in general, appear curved?

So could I say that if our stationary observer $O$ setup Riemann normal coordinates at $E$, it would not describe his "rest frame", unlike Fermi normal coordinates along his worldline, but rather the "rest frame" of a freely falling observer $O'$ intersecting his worldline at $E$? Would this $O'$ necessarily have to be comoving with $O$ at $E$ if this interpretation if Riemann normal coordinates is not horribly flawed?

Finally, even in light of the previous paragraph, the fact that Riemann normal coordinates physically has something special for freely falling observers does nothing to prevent the fact that if $O$ sets them up at $E$, while hovering at $r$, he will still be able to write down non-gravitational tensor laws evaluated at $E$ in SR form due to Christoffel symbols vanishing at that event in these coordinates correct?

Thank you so much for all your help Peter! You're awesome Sorry for all the questions xD

 

[Ben Niehoff]

This is true, but there's another difference, which I just alluded to. Let's take a concrete example: the worldline of an observer "hovering" at constant r in Schwarzschild coordinates. This observer is accelerated, so we can set up Fermi normal coordinates centered on his worldline. In these coordinates, the observer's worldline is a straight line.

Now pick an event E on that observer's worldline, and consider the worldline of a freely falling observer who just comes to rest for an instant at radial coordinate r at that event E. In the above Fermi normal coordinates (let's suppose we pick event E as the origin of those coordinates, to make it simple), the free-faller's worldline will be curved (I think it will look something like the way an inertial observer's worldline in Minkowski spacetime looks in Rindler coordinates). But if we set up Riemann normal coordinates centered on event E, in those coordinates the free-faller's worldline is a straight line, and the "hovering" observer's worldline will be curved (I think it will be a hyperbola, as a Rindler observer's worldline is in Minkowski coordinates in flat spacetime).

I'm not sure in what sense any of this is relevant. What you are really saying is that "If I take some coordinates and then draw a coordinate chart where each of my coordinates is plotted on a Cartesian axis, then the lines of constant coordinate values look like straight lines in my chart". So what? How does such a chart relate to anything physical? This is no different than saying "If I plot the radius $r$ and the angle $\theta$ on Cartesian axes, then a circle looks like a straight line".

You seem to be implying that each observer has a right to say "I am at rest and that other guy is on a curved worldline," but this is not so. The observers are not equivalent. The accelerating observer has a proper acceleration that he can measure in his own frame, and thereby conclude that he is the one in non-uniform motion.

For example, right at this moment, you are a hovering observer in a Schwarzschild geometry. And you can feel the ground pushing on your feet. The equivalence principle states that you can locally consider yourself to be stationary in a constant Newtonian gravitational field, but this is not the same as being able to consider yourself inertial.

Local flatness is not an observer-dependent notion; it is a basic axiom of (pseudo)-Riemannian geometry. Therefore every observer is capable of considering locally inertial frames; and observers can certainly measure, against such a frame, whether their worldline is locally geodesic or not.

 

[PAllen]

Consider Rindler coordinates in flat spacetime. These coordinates make the worldlines of accelerated observers look like straight lines. Question: are Rindler coordinates equivalent to Fermi-Normal coordinates for one of those accelerated observers?

I have been confused by this in the past. Take Rindlier coordinates as normally presented (for the whole family of accelerated observers), with origin on the horizon, and perform a simple translation coordinate change so as to place the origin on a particular observer, and you get FN coordinates for that specific uniformly accelerating observer. Which observer you translate to, to get FN, depends on the constant g used in the Rindler metric.

 

[PeterDonis]

I'm not sure in what sense any of this is relevant.

It doesn't change any of the physics, true; it certainly doesn't change which worldlines are geodesics and which aren't, or make an inertial observer equivalent to an accelerated one, I agree. But it does illustrate a difference in how the different kinds of coordinate charts describe the physics.

 

[PeterDonis]

the locally flat coordinate system is the system of a freely falling observer at the point in question of the space-time."

I think what he means by this is that the "time axis" of the locally flat coordinate system will be the worldline of some inertial observer, as opposed to some accelerated observer, i.e., an observer "at rest" in this coordinate system will be inertial.

for an accelerating observer, at $P_0$, I can interpret this to mean we take a momentarily comoving locally inertial observer at $P_0$ with the associated frame $\{e_{\alpha}(P_0)\}$?

That's how I read it, yes.

So are Fermi Normal coordinates sensitive to the worldline we choose them to be along?

Yes. Once again, to summarize the difference between F-N and R-N coordinates:

(1) R-N coordinates are chosen to make the metric coefficients Minkowski *and* the connection coefficients (i.e., the first derivatives of the metric coefficients) all zero *at a particular event*.

Note that this does not mean there is one unique set of R-N coordinates at a particular event E. There is a 6-parameter group of them, corresponding to the 6 parameters of local Lorentz transformations that keep the origin at E (or, equivalently, the 6 parameters describing the set of possible inertial worldlines passing through E).

(2) F-N coordinates are chosen to make the metric coefficients Minkowksi *all along a particular worldline*.

This means that, for a given worldline, there is one unique set of F-N coordinates for it (leaving out trivial transformations like rescaling the units). In these coordinates, the chosen worldline will be the "time axis". Other different worldlines, even if they share events with the chosen worldline, will look different.

In this second case wouldn't Fermi normal coordinates evaluated at E just be Riemann normal coordinates at E?

Meaning, if the worldline we choose for F-N coordinates is a geodesic passing through E? I think so, yes.

So in a sense, can we consider Riemann normal coordinates to describe unequivocally the "rest frame" of freely falling observers

Only if you recognize that there is not a unique such "rest frame", but a 6-parameter group of them, as above. But at a given event, in a given spacetime, that 6-parameter group is uniquely defined; it doesn't depend on picking a worldline or anything else, it depends only on the geometry of the spacetime at the event we picked.

So could I say that if our stationary observer $O$ setup Riemann normal coordinates at $E$, it would not describe his "rest frame", unlike Fermi normal coordinates along his worldline, but rather the "rest frame" of a freely falling observer $O'$ intersecting his worldline at $E$? Would this $O'$ necessarily have to be comoving with $O$ at $E$ if this interpretation if Riemann normal coordinates is not horribly flawed?

Yes, this looks correct. Put another way, if two worldlines have the same tangent vector at E, then they will both have the same "natural" R-N coordinates at E, because a given tangent vector at E determines a unique geodesic, and thereby picks out a unique set of R-N coordinates from the 6-parameter group of possible ones at E.

Finally, even in light of the previous paragraph, the fact that Riemann normal coordinates physically has something special for freely falling observers does nothing to prevent the fact that if $O$ sets them up at $E$, while hovering at $r$, he will still be able to write down non-gravitational tensor laws evaluated at $E$ in SR form due to Christoffel symbols vanishing at that event in these coordinates correct?

Yes.

 

[WannabeNewton]

Yes, this looks correct. Put another way, if two worldlines have the same tangent vector at E, then they will both have the same "natural" R-N coordinates at E, because a given tangent vector at E determines a unique geodesic, and thereby picks out a unique set of R-N coordinates from the 6-parameter group of possible ones at E.

Thank you so much Peter. I just have a question about this particular point. I'm not understanding what you mean by the two comoving observers at E will have the same RN coordinates. Same in what sense? How would the RN coordinates setup by the two comoving observers at E differ from the RN coordinates setup by an observer at E who wasn't comoving with them?

Do you mean like, the two comoving observers can naturally choose the same momentarily comoving LIF at E so that the orthonormal basis chosen to setup the RN coordinates, coming from the comoving LIF at E, will be the same for both? This would then result in the same RN coordinate chart. The nocomoving observer at E would end up with a different RN coordinate chart if he took the orthonormal basis associated with his momentarily comoving LIF so is that what you meant by "natural" RN chart i.e. the RN chart coming from the orthonormal basis of the momentarily comoving LIF? Because the noncomoving observer could also apply infinitesimal Lorentz transformations until his orthonormal basis coincides with that of the two comoving observers can't he? Then he would get the same RN chart as then. This, however, won't be a "natural" choice of RN chart for the noncomoving observer in the above sense.

Finally, I would just like to address once and for all the statements made in some of my GR texts that brought on this entire confusion in the first place. For example, in page 173 of Schutz, the author states the Einstein Equivalence principle as follows: "Any local physical experiment not involving gravity will have the same result if performed in a freely falling inertial frame as if it were performed in the flat spacetime of special relativity."

Now as we have just discussed, any observer (accelerating or not) can setup RN coordinates about an event and write down nongravitational GR laws as regular SR laws at that event. Schutz, on the other hand, is identifying RN coordinates about an event with a freely falling inertial frame at that event. Why exactly can we make this identification?

This is the language I see in many texts that thoroughly confuses me. Is it again the statement that at that event there exists a freely falling observer for whom the associated worldline at that event is along the "time" axis of the RN coordinates (or rather the "time axis" of the original orthonormal frame chosen when setting up the RN coordinates) so we can imagine the RN coordinates as describing the "rest frame" of some freely falling observer at that event and so just call the RN coordinates about the event a freely falling frame since it essentially acts as the reference frame of some freely falling observer at that event?

Thanks again!

 

[PeterDonis]

How would the RN coordinates setup by the two comoving observers at E differ from the RN coordinates setup by an observer at E who wasn't comoving with them?

I should have clarified that I meant R-N coordinates in which the comoving observers are at rest at E. Obviously if you don't require an observer to be at rest in the R-N coordinates at E, any observer can use any R-N coordinates at E that they like. But any given observer will only be at rest at E in one particular set of R-N coordinates at E, the one determined by their tangent vector at E.

The nocomoving observer at E would end up with a different RN coordinate chart if he took the orthonormal basis associated with his momentarily comoving LIF so is that what you meant by "natural" RN chart i.e. the RN chart coming from the orthonormal basis of the momentarily comoving LIF?

Yes.

Because the noncomoving observer could also apply infinitesimal Lorentz transformations until his orthonormal basis coincides with that of the two comoving observers can't he?

Yes, but as you note, the new basis won't be "natural" for him. I was talking specifically about the "natural" R-N chart for a given observer at E.

the author states the Einstein Equivalence principle as follows: "Any local physical experiment not involving gravity will have the same result if performed in a freely falling inertial frame as if it were performed in the flat spacetime of special relativity."

Now as we have just discussed, any observer (accelerating or not) can setup RN coordinates about an event and write down nongravitational GR laws as regular SR laws at that event. Schutz, on the other hand, is identifying RN coordinates about an event with a freely falling inertial frame at that event. Why exactly can we make this identification?

I think there is a common vagueness of terminology in many relativity texts surrounding the word "frame". Strictly speaking, a "freely falling frame" does not refer to a coordinate chart; it refers to an orthonormal frame carried by a freely falling observer. So the statement of the EEP you quoted, as I read it, isn't really a statement about coordinate charts, R-N or otherwise; it's a statement about the actual physical measurements that a freely falling observer makes.

However, this brings up another point, which I think your questions are getting at: on the interpretation I just gave, the statement of the EEP you quoted is too weak! After all, consider an accelerating observer who is momentarily comoving with a freely falling observer at event E. The physical measurements of the accelerating observer at E will be the same as those of an accelerating observer in flat spacetime, just as the physical measurements of the freely falling observer at E will be the same as those of an inertial observer in flat spacetime.

Now one could try to finesse this by saying that, since the two observers (accelerated and free-falling) are momentarily comoving at E, the tangent vectors of their worldlines are identical at E, and therefore they both have the same orthonormal basis at E, and hence the same "frame" at E. The problem with this is that, even if they have the same "frame" at E, the "rate of change" of that frame is different for the two of them, and many (if not most) physical experiments really depend on rates of change, not just values at a single event. (The simplest example would be the readings on accelerometers carried by each observer; one will be zero and one nonzero, but this difference is purely one of rate of change of the respective tangent vectors at E.)

The only way to include rates of change is, of course, to expand one's "frame" to include, not just the single event E, but a small patch of spacetime surrounding it. But then what we are really deailng with is a frame field, not a frame--and the frame fields associated with the two observers (accelerated and inertial) will be *different*. The "freely falling frame" on this view can only mean the frame field associated with the inertial observer; but again, that makes the statement of the EEP as given too weak. To really state it correctly you have to be more precise about both the physics and the terminology than most texts are that I've seen.

 

[WannabeNewton]

I should have clarified that I meant R-N coordinates in which the comoving observers are at rest at E. Obviously if you don't require an observer to be at rest in the R-N coordinates at E, any observer can use any R-N coordinates at E that they like. But any given observer will only be at rest at E in one particular set of R-N coordinates at E, the one determined by their tangent vector at E.

And this is just the RN chart coming from the choice of orthonormal basis / frame in which $e_0(E) = u(E)$, where $u(E)$ is the 4-velocity of the observer at $E$, right? And this will physically correspond to picking the frame of a momentarily comoving freely falling observer at $E$.

I think there is a common vagueness of terminology in many relativity texts surrounding the word "frame". Strictly speaking, a "freely falling frame" does not refer to a coordinate chart; it refers to an orthonormal frame carried by a freely falling observer. So the statement of the EEP you quoted, as I read it, isn't really a statement about coordinate charts, R-N or otherwise; it's a statement about the actual physical measurements that a freely falling observer makes.

Certainly, which is why I was confused because the author (Schutz) uses the word "free falling locally inertial frame" and "locally flat coordinates" synonymously throughout the text

Now one could try to finesse this by saying that, since the two observers (accelerated and free-falling) are momentarily comoving at E, the tangent vectors of their worldlines are identical at E, and therefore they both have the same orthonormal basis at E, and hence the same "frame" at E. The problem with this is that, even if they have the same "frame" at E, the "rate of change" of that frame is different for the two of them, and many (if not most) physical experiments really depend on rates of change, not just values at a single event. (The simplest example would be the readings on accelerometers carried by each observer; one will be zero and one nonzero, but this difference is purely one of rate of change of the respective tangent vectors at E.)

Right so even if $e_0(E) = u(E)$ is the same for both the accelerating and freely falling observer, $\nabla_{e_0}e_0|_E \neq 0$ for the accelerating observer even though $\nabla_{e_0}e_0|_E = 0$ for the freely falling one since the accelerating observer will still be able to see a non-zero reading on his accelerometer regardless of if he adopts momentarily comoving LIF of a freely falling observer at $E$.

Ah so is this what you meant by the quoted EEP being too weak? Both the accelerating and freely falling observers have the same freely falling frame at $E$ but as discussed there are many physical experiments the accelerating observer can do at $E$ that will yield different results from identical ones performed by the freely falling observer (e.g. the accelerometer reading) so it is not enough to just say any experiments performed in a freely falling frame at $E$ will yield the same results as identical experiments performed in SR because there is an ambiguity here. It makes it seem like even an accelerating observer who adopts a momentarily comoving freely falling frame at $E$ will get SR results identical to those of a freely falling observer at $E$ when in reality he just gets SR results identical to those of the same accelerating observer in SR. Correct?

Also, just a question on terminology: is a freely falling frame at $E$ the same as a locally inertial frame at $E$ since both correspond to some freely falling observer at $E$ (who is by definition a locally inertial observer)? Both would just correspond to an orthonormal basis carried by a freely falling (i.e. locally inertial) observer at $E$ such that $e_0(E) = u(E)$ right?

The only way to include rates of change is, of course, to expand one's "frame" to include, not just the single event E, but a small patch of spacetime surrounding it. But then what we are really deailng with is a frame field, not a frame--and the frame fields associated with the two observers (accelerated and inertial) will be *different*. The "freely falling frame" on this view can only mean the frame field associated with the inertial observer; but again, that makes the statement of the EEP as given too weak. To really state it correctly you have to be more precise about both the physics and the terminology than most texts are that I've seen.

The two frame fields will be different in the small patch of space-time because for the accelerating observer we will need a different freely falling frame at each event along his worldline contained in the patch whereas for the freely falling observer we can trivially have the same freely falling frame at each point along his worldline within the patch since this frame would get parallel transported along the worldline right?

So is this why, if we are to think of the "freely falling frame" as really describing a frame field, we can only make sense of it for the freely falling observer since for the accelerating observer we won't have a single "freely falling frame" in the patch, which would make the concept of ascribing a "freely falling frame" to a frame field associated with the accelerating observer ill defined? In what way would the EEP then be too weak with regards to this? Would it then be unable to state the equivalence of local experiments performed by accelerating observers in space-time with identical experiments performed by the same accelerating observers in SR (which we know the EEP does state, event by event)?

Thanks!

 

[PeterDonis]

It makes it seem like even an accelerating observer who adopts a momentarily comoving freely falling frame at $E$ will get SR results identical to those of a freely falling observer at $E$ when in reality he just gets SR results identical to those of the same accelerating observer in SR. Correct?

Yes.

The two frame fields will be different in the small patch of space-time because for the accelerating observer we will need a different freely falling frame at each event along his worldline contained in the patch whereas for the freely falling observer we can trivially have the same freely falling frame at each point along his worldline within the patch since this frame would get parallel transported along the worldline right?

Yes, although I would say it's even simpler: the frames carried by each observer can only match up at a single event, because one is accelerated and the other is inertial.

if we are to think of the "freely falling frame" as really describing a frame field, we can only make sense of it for the freely falling observer since for the accelerating observer we won't have a single "freely falling frame" in the patch, which would make the concept of ascribing a "freely falling frame" to a frame field associated with the accelerating observer ill defined?

Yes.

In what way would the EEP then be too weak with regards to this? Would it then be unable to state the equivalence of local experiments performed by accelerating observers in space-time with identical experiments performed by the same accelerating observers in SR?

If you use the term "freely falling frame" in stating the EEP, then it seems obvious, given the above, that it can only apply to measurements made by a freely falling observer, yes?

(which we know the EEP does state, event by event)

We "know" this because we intuitively extend the EEP to accelerated observers, since that's obviously the intent. But stating the EEP using the term "freely falling frame" does not, strictly speaking, justify this extension of its scope (at least IMO).

 

[WannabeNewton]

If you use the term "freely falling frame" in stating the EEP, then it seems obvious, given the above, that it can only apply to measurements made by a freely falling observer, yes?
...
We "know" this because we intuitively extend the EEP to accelerated observers, since that's obviously the intent. But stating the EEP using the term "freely falling frame" does not, strictly speaking, justify this extension of its scope (at least IMO).

So in light of these two statements and everything discussed above, should we just be cautious with the way many textbooks phrase this part of the EEP, that is when they use the term "freely falling frame" (because I see it a lot)?

The full power of the EEP does allow an accelerating observer to use SR locally in the manner discussed amongst the previous posts i.e. via RN coordinates setup by the accelerating observer at an event, using some orthonormal frame carried by a momentarily comoving freely falling observer (and the orthonormal frame may or may not correspond to the rest frame of the comoving observers at that event). There is no contradiction because as you mentioned the accelerating observers will not get the same results from local experiments as purely freely falling observers, even if they are comoving at the event in question.

I think this is why the way in which many textbooks state the EEP, i.e. in such an ambiguous way using the term "freely falling frame", manages to confuse me and the OP. It makes it seem like the EEP / local flatness about an event is somehow restricted exclusively to freely falling observers even though it is a local property of space-time itself (honestly I think this is the easiest way to state it).

Interestingly, Wald never once uses these confusing terms in the text. He doesn't even mention the terms locally inertial coordinates and locally inertial / freely falling frames once in the text. This can be a good thing if you want to avoid the confusions you could potentially get due to the way such things are presented in other texts but it is also a bad thing simply because you will never gain experience with those concepts if you only use Wald (which was where I got screwed over ).

 

[PeterDonis]

should we just be cautious with the way many textbooks phrase this part of the EEP, that is when they use the term "freely falling frame" (because I see it a lot)?

I think that would help, yes.

 

[WannabeNewton]

I think that would help, yes.

Thank you so much for all the help Peter. This discussion has been really enlightening; it has cleared up many of the confusions I have acquired from the various texts mentioned (except Wald xD) and hopefully for the OP as well. Until next time

 

[BruceW]

so, specifically to the OP's question, he wanted to know if a non-free-falling observer could measure the energy of a photon (or beam of light really) and would this give the same answer as for an observer who was free-falling. (if I have interpreted it correctly). From what I've seen, I think the answer is yes. This is because even though there will be fictitious forces acting on the beam of light, we are not making any measurements that depend on its acceleration. Our measurement is of its frequency, so the acceleration of the beam of light does not matter. It will cause the frequency of the light beam to change as it passes the observer, but the observer can just choose the value of the light's frequency when it passes by closest to him.

And if the observer wanted to do an experiment that did involve accelerations, then he would get different results compared to the freely-falling observer. We just happen to be lucky that the experiment described by the OP'er doesn't depend on accelerations. Does this all sound about right?